Difference between revisions of "2019 AIME II Problems/Problem 9"

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==Solution 2==
 
==Solution 2==
For <math>n</math> to be 20-pretty, <math>n</math> can only take on certain prime factorization forms: namely, <math>p^{19}, p^9q, p^4q^3, p^4qr</math>. Notice that for the first form, the smallest possible integer with it is <math>2^{19}</math> and the smallest integer with the second form divisible by 20 is <math>2^{9}5</math> which are both greater than 2019.  
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For <math>n</math> to have exactly <math>20</math> positive divisors, <math>n</math> can only take on certain prime factorization forms: namely, <math>p^{19}, p^9q, p^4q^3, p^4qr</math>. No number that is a multiple of <math>20</math> can be expressed in the first form, and the only integer divisible by <math>20</math> that has the second form is <math>2^{9}5</math>, which is greater than <math>2019</math>.  
  
For the third form, the only numbers with it and divisible by 20 is <math>2^45^3=2000</math> and <math>2^35^4=5000</math> so only <math>2000</math> is 20-pretty with this factorization.  
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For the third form, the only <math>20</math>-pretty numbers are <math>2^45^3=2000</math> and <math>2^35^4=5000</math>, and only <math>2000</math> is small enough.  
  
For the fourth form, <math>2^45r</math> is the sufficient combination for <math>20|n</math>. Since <math>n=80r<2019</math>, <math>r\le 25</math>. Therefore, <math>r</math> can take on <math>3, 7, 11, 13, 17, 19,</math> or <math>23</math>.  
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For the fourth form, any number of the form <math>2^45r</math> where <math>r</math> is a prime other than <math>2</math> or <math>5</math> will satisfy the <math>20</math>-pretty requirement. Since <math>n=80r<2019</math>, <math>r\le 25</math>. Therefore, <math>r</math> can take on <math>3, 7, 11, 13, 17, 19,</math> or <math>23</math>.  
  
 
Thus, <math>\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}</math>.
 
Thus, <math>\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}</math>.

Revision as of 17:53, 4 June 2020

Problem 9

Call a positive integer $n$ $k$-pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$. For example, $18$ is $6$-pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$-pretty. Find $\tfrac{S}{20}$.

Solution

Every 20-pretty integer can be written in form $n = 2^a 5^b k$, where $a \ge 2$, $b \ge 1$, $\gcd(k,10) = 1$, and $d(n) = 20$, where $d(n)$ is the number of divisors of $n$. Thus, we have $20 = (a+1)(b+1)d(k)$, using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, there are not many cases to check.

If $a+1 = 4$, then $b+1 = 5$. But this leads to no solutions, as $(a,b) = (3,4)$ gives $2^3 5^4 > 2019$.

If $a+1 = 5$, then $b+1 = 2$ or $4$. The first case gives $n = 2^4 \cdot 5^1 \cdot p$ where $p$ is a prime other than 2 or 5. Thus we have $80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23$. The sum of all such $n$ is $80(3+7+11+13+17+19+23) = 7440$. In the second case $b+1 = 4$ and $d(k) = 1$, and there is one solution $n = 2^4 \cdot 5^3 = 2000$.

If $a+1 = 10$, then $b+1 = 2$, but this gives $2^9 \cdot 5^1 > 2019$. No other values for $a+1$ work.

Then we have $\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}$.

-scrabbler94

Solution 2

For $n$ to have exactly $20$ positive divisors, $n$ can only take on certain prime factorization forms: namely, $p^{19}, p^9q, p^4q^3, p^4qr$. No number that is a multiple of $20$ can be expressed in the first form, and the only integer divisible by $20$ that has the second form is $2^{9}5$, which is greater than $2019$.

For the third form, the only $20$-pretty numbers are $2^45^3=2000$ and $2^35^4=5000$, and only $2000$ is small enough.

For the fourth form, any number of the form $2^45r$ where $r$ is a prime other than $2$ or $5$ will satisfy the $20$-pretty requirement. Since $n=80r<2019$, $r\le 25$. Therefore, $r$ can take on $3, 7, 11, 13, 17, 19,$ or $23$.

Thus, $\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}$.

See Also

2019 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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